Roasting galena [lead(II) sulfide] is an early step in the industrial isolation of lead. How many liters of sulfur dioxide, measured at STP, are produced by the reaction of 3.73 kg of galena with 170. L of oxygen gas at 220°C and 2.00 atm? Lead(II) oxide also forms.

The volume of sulfur dioxide measured at STP is 125.44 L

Explanation:

To calculate the number of moles, we use the equation:

Given mass of galena = 3.73 kg = 3730 g    (Conversion factor:  1 kg = 1000 g)

Molar mass of galena = 239.3 g/mol

Putting values in above equation, we get:

To calculate the amount of oxygen gas, we use the equation given by ideal gas which follows:

where,

P = pressure of the gas = 2.00 atm  

V = Volume of the gas = 170 L

T = Temperature of the gas =

R = Gas constant =

n = number of moles of oxygen gas = ?

Putting values in above equation, we get:

The chemical equation for the reaction of lead (II) sulfide and oxygen gas follows:

By Stoichiometry of the reaction:

3 moles of oxygen gas reacts with 2 moles of galena

So, 8.4 moles of oxygen gas will react with = of galena

As, given amount of lead (II) sulfide is more than the required amount. So, it is considered as an excess reagent.

Thus, oxygen gas is considered as a limiting reagent because it limits the formation of product.

By Stoichiometry of the reaction:

3 moles of oxygen gas produces 2 moles of sulfur dioxide

So, 8.4 moles of oxygen gas will produce = of sulfur dioxide

At STP:

1 mole of a gas occupies 22.4 L of volume

So, 5.6 moles of sulfur dioxide will occupy = of volume

Hence, the volume of sulfur dioxide measured at STP is 125.44 L

radioactive decay occurs in unstable atomic nuclei – that is, ones that don't have enough binding energy to hold

the nucleus together due to an excess of either protons or neutrons.

"standard molar enthalpy:

h2o ( liquid water ) : - 285.8 kj/mole

h2o ( water vapor ) : - 241.8 kj / mole.

δhrxn = - 241. 8 - ( - 285.8 )= -241.8 + 285.8 = 44.0

answer: 44.0 kj"